# Homework Help: Nonuniform Circular Motion: Old-Fashioned Amusement Park Ride

1. Oct 22, 2012

### Go Boom Now

Sorry for having my first post be a question, I guess. I'm just confused as to how to do this since I've never really been one for word problems. I can't really type in... tex(?) either. Oh well, here we go:

1. The problem statement, all variables and given/known data
In an old-fashioned amusement park ride, passengers stand inside a 5.0-m-diameter hollow steel cylinder with their backs against the wall. The cylinder begins to rotate about a vertical axis. The the floor on which the passengers are standing on suddenly drops away! If all goes well, the passengers will "stick" to the wall and not slide. Clothing has a static coefficient of friction against steel in the range of 0.60 to 1.0 and kinetic coefficient of friction in the range of 0.40 to 0.70. A sign next to the entrance says "No children under 30kg allowed." What is the minimum angular speed, in RPM, for which the ride is safe?

2. Relevant equations
Kinetic Friction Force = Coefficient of Kinetic Friction x Normal Force
Static Friction Force = Coefficient of Static Friction x Normal force
F-net = mass x acceleration = (mass x (tangential velocity^2)/radius
Tangential Velocity = (2pi x radius)/period = angular velocity x radius

3. The attempt at a solution
Check the attachment. I was following the steps my teacher outlined (FBD in r, t, z components, net force equations, solve). I'm not sure if my diagrams or FBD are correct though because the people are also involved... do I just bundle them up in the steel cylinder?

Sorry, I'm just confused.

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2. Oct 23, 2012

### tiny-tim

Welcome to PF!

Hi Go Boom Now! Welcome to PF!
That's normal!
yes, centripetal acceleration = v2/r = ω2r (even if ω isn't constant)

your only body for a free body diagram is the person, so i don't understand what you mean about including them in the cylinder

your diagrams, and your z equation, look ok

your x equation (i think you know) is normal force = mass times centripetal acceleration (and that's where you start)

what is worrying you about that?

3. Oct 23, 2012

### Go Boom Now

I drew the wrong kind of diagram (I misinterpreted the question), which is why I ended up getting confused. I managed to figure it out after I read your post. Thanks for the help!